
\begin{answer}
\begin{enumerate}
    \item 
        For any $s$,
$$
\begin{aligned}
|B(V_1)(s) - B(V_2)(s)| &= \gamma |\max_{a\in A} \sum_{s'\in S}P_{sa}(s')V_1(s') - \max_{a\in A} \sum_{s' \in S}P_{sa}(s') V_2(s')|\\
&\le \gamma \max_{a\in A}|\sum_{s'\in S}P_{sa}(s')V_1(s') - \sum_{s' \in S}P_{sa}(s')V_2(s')|\\
&= \gamma \max_{a\in A}|\sum_{s'\in S}P_{sa}(s')(V_1(s') - V_2(s'))|\\
&\le \gamma \max_{a\in A}\sum_{s'\in S}P_{sa(s')}\|V_1 - V_2\|_{\infty}\\
&= \gamma \|V_1 -V_2\|_\infty \max_{a\in A}\sum_{s'\in S}P_{sa}(s')\\
&= \gamma \|V_1 - V_2\|_{\infty}
\end{aligned}
$$
Since this is true for any $s$, we get
$$
\|B(V_1) - B(V_2)\|_\infty \le \gamma \|V_1 - V_2\|_\infty
$$

\item 

    Suppose that $V_1$ and $V_2$ are two fixed points, then
$$
\|V_1 - V_2\|_\infty = \|B(V_1) - B(V_2)\|_\infty \le \gamma \|V_1 - V_2\|_\infty
$$
This is true only when $\|V_1 - V_2\|_\infty = 0$, or $V_1 = V_2$. So fixed point is unique.






\end{enumerate}
\end{answer}
